Problem: The following function gives the cost, in dollars, of producing $x$ kilograms of fertilizer: $C(x)=0.001x^3-0.14x^2+7x+160$ What is the instantaneous rate of change of the cost when $50$ kilograms are produced? Choose 1 answer: Choose 1 answer: (Choice A) A $-6.5$ kilograms per dollar (Choice B) B $-6.5$ dollars per kilogram (Choice C) C $0.5$ kilograms per dollar (Choice D) D $0.5$ dollars per kilogram
Solution: Understanding the problem The function that represents the instantaneous rate of change of $C(x)$ is its derivative, $C'(x)$. Therefore, the instantaneous rate of change of the cost when $50$ kilograms are produced is $C'(50)$. Let's find $C'(x)$ and evaluate it at $x=50$. Finding $C'(x)$ $C'(x)=0.003x^2-0.28x+7$ Finding $C'(50)$ $\begin{aligned} C'({50})&=0.003({50})^2-0.28({50})+7 \\\\ &=0.5 \end{aligned}$ Interpreting units $C(x)$ is the cost in ${\text{dollars}}$ for $x$ ${\text{kilograms}}$. Therefore, we measure its rate of change in ${\text{dollars}}$ per ${\text{kilogram}}$. In conclusion, the instantaneous rate of change of the cost when $50$ kilograms are produced is $0.5$ dollars per kilogram. The rate of change is positive because the cost is increasing.